Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Derivative of x^2:

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
|dw:1323757204190:dw|
|dw:1323757255000:dw|
y=x^2 dy/dx=2x we know derivative of x^n=nx^(n-1)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I'm not asking :) hold on
That tangent line has an equation \[ y - x_0^2 = m(x-x_0) \] for some m. Let x > x0. Then, since the points on the line are less than the points on the curve, \[ x^2 \geq m(x-x_0) + x_0^2\] \[x^2-x_0^2 = (x+x_0)(x-x_0) \geq m(x-x_0) \] and x-x0 is positive, so \[x+x_0 \geq m\] Next, for x < x0, the points on the line are likewise less than the points of the curve, so we get the same equation but now x-x0 < 0 so we obtain \[x+x_0 \leq m\] now, for the first one, since x must be greater than or equal to x0, we can say that \[x+x_0 \geq 2x_0 \geq m\] and Likewise for the second, since x must be less than x0, \[x + x_0 \leq 2x_0 \leq m\] thus \[2x_0 \leq m \leq 2x_0\] and thus \[m = 2x_0\] so the slope of the tangent line to the curve y=x^2 at some point x0 is equal to 2x0, no difference quotients or limits required. Ta da!
Don't judge me, I was bored and decided to try a derivative without a limit. Shush. :)
It's really interesting, actually.

Not the answer you are looking for?

Search for more explanations.

Ask your own question